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I have a closed subspace $Y$ of a Banach space $X$ and a map $T: X'/Y^{\circ} \to Y'$ given by $[f] \to f|_y$. The norm in $X'/Y^\circ$ is given by $\|[f]\| = \inf \{ \|f-h\| : h \in Y^\circ \}$. I'm trying to show this map is an isometry, i.e. $\inf \{ \|f-h\| : h \in Y^\circ \} = \inf \{M > 0 : |f(y)| \le M \|y\| \}$ I've shown it is well-defined, linear and I've shown one side of the isometry inequality:

If $h \in Y^{\circ}$ then $|f(y)| = |(f-h)(y)| \le \|f-h\|\|y\|$. Hence $||f|_y \le \|[f]\|$. Now for the other inequality I want to show that if I take an $M$ in the set $\{M > 0 : |f(y)| \le M \|y\| \}$ then this is the "best" bound for some particular $\|f-h\|$ and then we would have $\inf \{ \|f-h\| : h \in Y^{\circ} \} \ge \inf \{M > 0 : |f(y)| \le M \|y\| \}$.

To do this I can see for a general $h \in Y^{\circ}$ we have $|f(y)-h(y)| \le M\|y\|$ and so $\|f-h\| \le M$. Now I suppose $K$ is a bound for $f-h$ and hence $K \ge \frac{|f(x)-h(x)|}{\|x\|}$ and now I'd like to show $K \ge M$ which would then give $\|f-h\| = M$ but I'm unsure how to pick $h$ to achieve this!

Apologies for the lengthy explanation, I hope this is okay to follow.

Michael Hardy
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Wooster
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    Notice that $\displaystyle||f-h||||y||$ looks different from $\displaystyle|f-h||y|$. The latter is standard and I changed it. ${}\qquad{}$ – Michael Hardy Jan 13 '15 at 14:56
  • I just noticed that in the edit, I will remember that, thanks! – Wooster Jan 13 '15 at 14:57
  • Maybe the following can help you: http://math.stackexchange.com/questions/1095671/question-about-a-proof-in-rudins-book-annihilators/1102203 – Janko Bracic Jan 14 '15 at 05:08

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